We know log_add, but how to do log_subtract?

Question

Multiplying two numbers in log space means adding them:

log_multiply(x, y) = log( exp(x) * exp(y) )
                   = x + y

Adding two numbers in log space means you do a special log-add operation:

log_add(x, y) = log( exp(x) + exp(y) )

which is implemented in the following code, in a way that doesn't require us to take the two exponentials (and lose runtime speed and precision):

  double log_add(double x, double y) {
    if(x == neginf)
      return y;
    if(y == neginf)
      return x;
    return max(x, y) + log1p(exp( -fabs(x - y) ));
  }

(Here is another one.)

But here is the question:

Is there a trick to do it for subtraction as well?

log_subtract(x, y) = log( exp(x) - exp(y) )

without having to take the exponents and lose precision?

  double log_subtract(double x, double y) {
    // ?
  }

Answer 1

How about

double log_subtract(double x, double y) {
  if(x <= y)
    // error!! computing the log of a negative number
  if(y == neginf)
    return x;
  return x + log1p(-exp(y-x));
}

That's just based on some quick math I did...